# What is the derivative of h(x)=\sin2x\cos2x

What is the derivative of $$\displaystyle{h}{\left({x}\right)}={\sin{{2}}}{x}{\cos{{2}}}{x}$$

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Orlando Paz
Step 1
Product Rule of differentiation:
If f(x) and g(x) are any two differentiable functions, then
$$\displaystyle{f{{\left({g}\right)}}}'{\left({x}\right)}={f{{\left({x}\right)}}}\times{g}'{\left({x}\right)}+{f}'{\left({x}\right)}\times{g{{\left({x}\right)}}}$$
Chain rule of differentiation:
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{f{{\left({g{{\left({x}\right)}}}\right)}}}={f}'{\left({g{{\left({x}\right)}}}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{g{{\left({x}\right)}}}$$
$$\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}$$
$$\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}$$
The given function is $$\displaystyle{h}{\left({x}\right)}={\sin{{2}}}{x}{\cos{{2}}}{x}$$
Step 2
Differentiate the given function using the product rule as follows.
$$\displaystyle{h}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}{\cos{{2}}}{x}\right)}$$
$$\displaystyle={\sin{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\cos{{2}}}{x}\right)}+{\cos{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}\right)}$$
(U sin g product rule)
$$\displaystyle={\sin{{2}}}{x}{\left(-{\sin{{2}}}{x}\right)}{\left({2}\right)}+{\cos{{2}}}{x}{\left({\cos{{2}}}{x}\right)}{\left({2}\right)}$$
(U sin g chain rule)
$$\displaystyle=-{2}{{\sin}^{{{2}}}{2}}{x}+{2}{{\cos}^{{{2}}}{2}}{x}$$
$$\displaystyle=-{2}{\left({1}-{{\cos}^{{{2}}}{2}}{x}\right)}+{2}{{\cos}^{{{2}}}{2}}{x}$$
$$\displaystyle=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}$$
Therefore, the derivative of the given function is
$$\displaystyle{h}'{\left({x}\right)}=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}$$
###### Not exactly what you’re looking for?
peterpan7117i
Step 1
Given function:
$$\displaystyle{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}{\left({2}{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}\right)}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}{\sin{{\left({4}{x}\right)}}}$$
Differentiating given function w.r.t x as follows
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{1}}}{{{2}}}}{\sin{{\left({4}{x}\right)}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{\left({4}{x}\right)}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\cos{{\left({4}{x}\right)}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({4}{x}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\cos{{\left({4}{x}\right)}}}{\left({4}\right)}$$
$$\displaystyle={2}{\cos{{\left({4}{x}\right)}}}$$
karton

Expalnation:
differentiate using the $$\frac{\text{product}}{\text{chain rule}}$$
$$\begin{array}{} \text{Given} \\y=f(x)g(x) \\\frac{dy}{dx}=f(x)g'(x)+g(x)f'(x)\Rightarrow \text{product rule} \\f(x)=\sin2x\Rightarrow f'(x)=2\cos2x \\g(x)=\cos2x\Rightarrow g'(x)=-2\sin2x \\\frac{d}{dx}(\sin2x\cos2x) \\=-2\sin^{2}2x+2\cos^{2}2x \\=2(\cos^{2}2x-\sin^{2}2x)=2\cos4x \end{array}$$